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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #22 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-10}\frac{(936x + 241x^{2} + 26x^{3} + x^{4} + 1260)}{(15x + x^{2} + 50)}\end{align*}$

Possible Answers:

$-\infty$

$\infty$

$\frac{ 84}{5}$

Correct answer:

$\frac{ 84}{5}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=-10\text{, as that would}\\&\text{create a zero value in the denominator. However, note that the expression}\\&\frac{(936x + 241x^{2} + 26x^{3} + x^{4} + 1260)}{(15x + x^{2} + 50)}\\&\text{can be factored. This will allow us to find the limit:}\\&\frac{((x + 3)\cdot (x + 6)\cdot (x + 7)\cdot (x + 10))}{((x + 5)\cdot (x + 10))}\\&\frac{((x + 3)\cdot (x + 6)\cdot (x + 7))}{(x + 5)}\\&lim_{x\rightarrow-10}\frac{(936x + 241x^{2} + 26x^{3} + x^{4} + 1260)}{(15x + x^{2} + 50)}=\frac{ 84}{5}\end{align*}$

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Example Question #981 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow2}\frac{(x - 2)}{(2x + x^{2} - 8)}\end{align*}$

Possible Answers:

$-\frac{1}{6}$

$\infty$

$-\infty$

$\frac{ 1}{6}$

Correct answer:

$\frac{ 1}{6}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=2\text{, as that would}\\&\text{create a zero value in the denominator. However, note that the expression}\\&\frac{(x - 2)}{(2x + x^{2} - 8)}\\&\text{can be factored. This will allow us to find the limit:}\\&\frac{(x - 2)}{((x - 2)\cdot (x + 4))}\\&\frac{1}{(x + 4)}\\&lim_{x\rightarrow2}\frac{(x - 2)}{(2x + x^{2} - 8)}=\frac{ 1}{6}\end{align*}$

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Example Question #24 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-2}\frac{(48x + 14x^{2} - 3x^{3} - x^{4} + 32)}{(461x + 67x^{2} - 5x^{3} - x^{4} + 630)}\end{align*}$

Possible Answers:

$-\infty$

$\infty$

$-\frac{4}{55}$

Correct answer:

$-\frac{4}{55}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=-2\text{, as that would}\\&\text{create a zero value in the denominator. However, note that the expression}\\&\frac{(48x + 14x^{2} - 3x^{3} - x^{4} + 32)}{(461x + 67x^{2} - 5x^{3} - x^{4} + 630)}\\&\text{can be factored. This will allow us to find the limit:}\\&\frac{((x + 1)\cdot (x + 2)\cdot (x - 4)\cdot (x + 4))}{((x + 2)\cdot (x + 5)\cdot (x + 7)\cdot (x - 9))}\\&\frac{((x + 1)\cdot (x - 4)\cdot (x + 4))}{((x + 5)\cdot (x + 7)\cdot (x - 9))}\\&lim_{x\rightarrow-2}\frac{(48x + 14x^{2} - 3x^{3} - x^{4} + 32)}{(461x + 67x^{2} - 5x^{3} - x^{4} + 630)}= -\frac{4}{55}\end{align*}$

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Example Question #23 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow8}\frac{(131x - 20x^{2} + x^{3} - 280)}{(68x - 15x^{2} + x^{3} - 96)}\end{align*}$

Possible Answers:

$-\infty$

$\infty$

$\frac{ 3}{20}$

Correct answer:

$\frac{ 3}{20}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=8\text{, as that would}\\&\text{create a zero value in the denominator. However, note that the expression}\\&\frac{(131x - 20x^{2} + x^{3} - 280)}{(68x - 15x^{2} + x^{3} - 96)}\\&\text{can be factored. This will allow us to find the limit:}\\&\frac{((x - 5)\cdot (x - 7)\cdot (x - 8))}{((x - 3)\cdot (x - 4)\cdot (x - 8))}\\&\frac{((x - 5)\cdot (x - 7))}{((x - 3)\cdot (x - 4))}\\&lim_{x\rightarrow8}\frac{(131x - 20x^{2} + x^{3} - 280)}{(68x - 15x^{2} + x^{3} - 96)}=\frac{ 3}{20}\end{align*}$

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Example Question #982 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow4}\frac{(x - 4)}{(x^{2} - 16)}\end{align*}$

Possible Answers:

$\frac{ 1}{8}$

$-\frac{1}{4}$

$-\infty$

$\infty$

Correct answer:

$\frac{ 1}{8}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=4\text{, as that would}\\&\text{create a zero value in the denominator. However, note that the expression}\\&\frac{(x - 4)}{(x^{2} - 16)}\\&\text{can be factored. This will allow us to find the limit:}\\&\frac{(x - 4)}{((x - 4)\cdot (x + 4))}\\&\frac{1}{(x + 4)}\\&lim_{x\rightarrow4}\frac{(x - 4)}{(x^{2} - 16)}=\frac{ 1}{8}\end{align*}$

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Example Question #983 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow7}\frac{(41x - 12x^{2} + x^{3} - 42)}{(x - 7)}\end{align*}$

Possible Answers:

$-\infty$

$\infty$

Correct answer:

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=7\text{, as that would}\\&\text{create a zero value in the denominator. However, note that the expression}\\&\frac{(41x - 12x^{2} + x^{3} - 42)}{(x - 7)}\\&\text{can be factored. This will allow us to find the limit:}\\&\frac{((x - 2)\cdot (x - 3)\cdot (x - 7))}{(x - 7)}\\&(x - 2)\cdot (x - 3)\\&lim_{x\rightarrow7}\frac{(41x - 12x^{2} + x^{3} - 42)}{(x - 7)}= 20\end{align*}$

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Example Question #28 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-3}\frac{(87x - 4x^{2} - x^{3} + 270)}{(37x - x^{3} + 84)}\end{align*}$

Possible Answers:

$\frac{ 42}{5}$

$\infty$

$-\infty$

Correct answer:

$\frac{ 42}{5}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=-3\text{, as that would}\\&\text{create a zero value in the denominator. However, note that the expression}\\&\frac{(87x - 4x^{2} - x^{3} + 270)}{(37x - x^{3} + 84)}\\&\text{can be factored. This will allow us to find the limit:}\\&\frac{((x + 3)\cdot (x - 9)\cdot (x + 10))}{((x + 3)\cdot (x + 4)\cdot (x - 7))}\\&\frac{((x - 9)\cdot (x + 10))}{((x + 4)\cdot (x - 7))}\\&lim_{x\rightarrow-3}\frac{(87x - 4x^{2} - x^{3} + 270)}{(37x - x^{3} + 84)}=\frac{ 42}{5}\end{align*}$

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Example Question #984 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-10}\frac{(3x + x^{2} - 70)}{(18x + x^{2} + 80)}\end{align*}$

Possible Answers:

$-\infty$

$\frac{ 17}{2}$

$\infty$

Correct answer:

$\frac{ 17}{2}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=-10\text{, as that would}\\&\text{create a zero value in the denominator. However, note that the expression}\\&\frac{(3x + x^{2} - 70)}{(18x + x^{2} + 80)}\\&\text{can be factored. This will allow us to find the limit:}\\&\frac{((x - 7)\cdot (x + 10))}{((x + 8)\cdot (x + 10))}\\&\frac{(x - 7)}{(x + 8)}\\&lim_{x\rightarrow-10}\frac{(3x + x^{2} - 70)}{(18x + x^{2} + 80)}=\frac{ 17}{2}\end{align*}$

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Example Question #26 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow7}-\frac{(x - 7)}{(49x + 4x^{2} - x^{3} - 196)}\end{align*}$

Possible Answers:

$-\frac{ 1}{42}$

$\infty$

$\frac{ 1}{42}$

$-\infty$

Correct answer:

$\frac{ 1}{42}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=7\text{, as that would}\\&\text{create a zero value in the denominator. However, note that the expression}\\&-\frac{(x - 7)}{(49x + 4x^{2} - x^{3} - 196)}\\&\text{can be factored. This will allow us to find the limit:}\\&\frac{(x - 7)}{((x - 4)\cdot (x - 7)\cdot (x + 7))}\\&\frac{1}{((x - 4)\cdot (x + 7))}\\&lim_{x\rightarrow7}-\frac{(x - 7)}{(49x + 4x^{2} - x^{3} - 196)}=\frac{ 1}{42}\end{align*}$

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Example Question #31 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-10}-\frac{(x + 10)}{(28x - 12x^{2} - x^{3} + 480)}\end{align*}$

Possible Answers:

$-\frac{ 1}{32}$

$\frac{ 1}{32}$

$-\infty$

$\infty$

Correct answer:

$\frac{ 1}{32}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=-10\text{, as that would}\\&\text{create a zero value in the denominator. However, note that the expression}\\&-\frac{(x + 10)}{(28x - 12x^{2} - x^{3} + 480)}\\&\text{can be factored. This will allow us to find the limit:}\\&\frac{(x + 10)}{((x - 6)\cdot (x + 8)\cdot (x + 10))}\\&\frac{1}{((x - 6)\cdot (x + 8))}\\&lim_{x\rightarrow-10}-\frac{(x + 10)}{(28x - 12x^{2} - x^{3} + 480)}=\frac{ 1}{32}\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #22 : Limits

Example Question #981 : Calculus 3

Example Question #24 : Limits

Example Question #23 : Partial Derivatives

Example Question #982 : Calculus 3

Example Question #983 : Calculus 3

Example Question #28 : Limits

Example Question #984 : Calculus 3

Example Question #26 : Partial Derivatives

Example Question #31 : Partial Derivatives

All Calculus 3 Resources

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